Question 199791
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Actually you were on the right track until you wrote:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_a = \frac{1}{\frac{1}{35}+\frac{1}{25}}]


Which is incorrect.  It should have been:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_a = \frac{2}{\frac{1}{35}+\frac{1}{25}}]


Since you were only taking out a factor of *[tex \Large D]


And then how you got from there to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_a = \frac{1\cdot\frac{60}{1}}{\frac{1}{60}\cdot\frac{60}{1}]


Is completely beyond me.  So lets go back to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_a = \frac{2}{\frac{1}{35}+\frac{1}{25}}]


The LCD is 175, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_a = \frac{2}{\frac{5}{175}+\frac{7}{175}} = \frac{2}{\frac{12}{175}} = 2\cdot\frac{175}{12} = \frac{175}{6} =29\frac{1}{6}]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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